01规划的模板题

链接：https://www.nowcoder.com/acm/contest/143/A
来源：牛客网

时间限制：C/C++ 1秒，其他语言2秒
空间限制：C/C++ 262144K，其他语言524288K
Special Judge, 64bit IO Format: %lld

题目描述

Kanade selected n courses in the university. The academic credit of the i-th course is s[i] and the score of the i-th course is c[i].

At the university where she attended, the final score of her is 

Now she can delete at most k courses and she want to know what the highest final score that can get.

输入描述:

The first line has two positive integers n,k

The second line has n positive integers s[i]

The third line has n positive integers c[i]

输出描述:

Output the highest final score, your answer is correct if and only if the absolute error with the standard answer is no more than 10-5

示例1

输入

3 1
1 2 3
3 2 1

输出

2.33333333333

说明

Delete the third course and the final score is 

备注:

1≤ n≤ 105

0≤ k < n

1≤ s[i],c[i] ≤ 103

题意：显而易见，这道题用到了01规划，通俗点讲就是有n种物品，每个物品都有a[i],b[i]两种不同的情况，选择n-k个元素要求

 最大，反观这道题，直接将数据带入即可

  我是使用的01规划中的二分法写的，但是这道题有一点要改进的就是至多删除k个元素，表示可以删除<=k个元素然后取最大值

这是我看的关于01规划算法的介绍https://blog.csdn.net/hzoi_ztx/article/details/54898323

主要公式：求max{f(r)}

#include <iostream>//二分法
#include <stdio.h>
#include <algorithm>
using namespace std;
int n,k,a[100005],b[100005];
double ps[100005];
const double eps=1e-5;
bool ok(double h)//f(r)=xi*a[i]*b[i]的和-xi*a[i]的和*r=(a[i]*b[i]的和-a[i]的和*r)*xi;为求最大值将xi全部为1，之后再将要删去的定为0.（xi={0,1}）
{
    for(int i=1;i<=n;i++) ps[i]=a[i]*b[i]-h*a[i];//ps数组表示横坐标为h是纵坐标的值后再将其全部加起来得到ans
    sort(ps+1,ps+1+n);
    double ans=0;                  //ans表示max{f(r)}即垂直于x轴坐标为h的直线与ps所代表的直线方程的交点；如果ans大于0则表示最大值在右边，否则在左边
    for(int i=n;i>=k+1;i--) ans+=ps[i];
    for(int i=2;i<=k;i++)
        if(ps[i]>0)ans+=ps[i];
    return ans>-eps;
}
void sol()
{
    for(int i=1;i<=n;i++) scanf("%d",a+i);
    for(int i=1;i<=n;i++) scanf("%d",b+i);
    double l=0,r=10000;
    while(r-l>eps)
    {
        double mid=(l+r)/2;
        if(ok(mid)) l=mid; else r=mid;
    }
    printf("%.11lf\n",l);
}
int main()
{
    scanf("%d%d",&n,&k);
    sol();
}